How many ordered pairs of digits (A, B) are possible in the number 479865AB so that the entire number is divisible by 9 and, in addition, the last digit B is odd?

Introduction / Context:
This problem combines the divisibility rule for 9 with conditions on the last digit of a number. We are given a seven digit number 479865AB, where A and B are digits from 0 to 9. The number must be divisible by 9, and the last digit B must be odd. We are asked to count how many ordered pairs (A, B) satisfy these conditions.

Given Data / Assumptions:
The number has the form 4 7 9 8 6 5 A B.Digits A and B can be any digits from 0 to 9.The number must be divisible by 9.The last digit B must be an odd digit.

Concept / Approach:
A number is divisible by 9 if and only if the sum of its digits is a multiple of 9. We first compute the sum of the known digits, then include A and B in that sum. We also restrict B to odd digits (1, 3, 5, 7, 9). Working modulo 9, we find all pairs (A, B) that make the total digit sum a multiple of 9, and then count how many such pairs are possible.

Step-by-Step Solution:
Step 1: Compute the sum of the fixed digits: 4 + 7 + 9 + 8 + 6 + 5 = 39.Step 2: The total sum of digits is 39 + A + B.Step 3: For divisibility by 9, we require 39 + A + B to be a multiple of 9.Step 4: Since 39 is already a multiple of 9 (39 = 9 × 4 + 3, so 39 ≡ 3 (mod 9)), we need A + B ≡ 6 (mod 9) so that 39 + A + B ≡ 3 + (A + B) ≡ 0 (mod 9).Step 5: The digits A and B must satisfy A + B = 6, 15 or 24, because these are the possible two digit sums in 0–9 range that are congruent to 6 modulo 9.Step 6: List all digit pairs (A, B) with A, B between 0 and 9 and B odd, satisfying these sums.Step 7: For A + B = 6 with B odd: possible pairs are (1, 5), (3, 3) and (5, 1).Step 8: For A + B = 15 with B odd: possible pairs are (6, 9), (8, 7). (Pairs like (7, 8) or (9, 6) are invalid because B must be odd and 8 or 6 are even.)Step 9: The sum 24 cannot be obtained with B odd and digits 0–9 simultaneously, because the only combinations involve at least one two digit or invalid digit.Step 10: Collecting valid pairs, we have (1, 5), (3, 3), (5, 1), (6, 9) and (8, 7), which is 5 pairs in total.

Verification / Alternative check:
We can verify by quickly checking the digit sum for each pair. For example, for (A, B) = (1, 5), the sum is 39 + 1 + 5 = 45, divisible by 9. For (6, 9), the sum is 39 + 6 + 9 = 54, also divisible by 9. All five pairs yield digit sums that are multiples of 9 and have B odd.

Why Other Options Are Wrong:
An answer of 6 or 9 overcounts and suggests the inclusion of invalid pairs where B is even or outside the 0–9 range. An answer of 11 grossly overcounts possible combinations and does not respect the digit constraints. Only 5 pairs meet all conditions.

Common Pitfalls:
Common mistakes include forgetting that B must be odd, miscomputing the sum of the fixed digits (39), or mis-applying modulo 9 arithmetic. Another frequent error is not considering all sums congruent to 6 modulo 9, or accidentally including pairs where A or B are not single digits.

Final Answer:
The number of ordered pairs (A, B) that satisfy all the conditions is 5.